Volume 2008, Article ID 395080,18pages doi:10.1155/2008/395080
Research Article
Multiplicity Results of Positive Radial Solutions for p -Laplacian Problems in Exterior Domains
Chan-Gyun Kim, Yong-Hoon Lee, and Inbo Sim
Department of Mathematics, Pusan National University, Pusan 609-735, South Korea
Correspondence should be addressed to Yong-Hoon Lee,[email protected] Received 23 December 2007; Accepted 16 March 2008
Recommended by Zhitao Zhang
We find the second positive radial solution for the followingp-Laplacian problem: div|∇u|p−2∇u K|x|uq 0 inΩ,u|∂Ω 0,ux→μ > 0 as|x| → ∞, whereΩ {x ∈ RN : |x| > r0},r0 > 0, N > p >1,K∈CΩ,0,∞andq > p−1. We also give some global existence results with respect to the parameterμ.
Copyrightq2008 Chan-Gyun Kim et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
In this paper, we study the existence, nonexistence, and multiplicity of positive radial solutions for the followingp-Laplacian problem:
div
|∇u|p−2∇u
K|x|uq 0 inΩ, P
u|∂Ω0, u−→μ >0 as|x| −→ ∞, D1
u|∂Ωμ >0, u−→0 as|x| −→ ∞, D2
whereΩ {x∈RN:|x|> r0>0}, N > p >1, μis a positive real parameter,K∈Cr0,∞,R withR 0,∞.
The present work is motivated by Deng and Li1who consider a semilinear problem of the form
ΔuKxuq0 inΩ, u >0 inΩ, u∈Hloc1 Ω∩C
Ω , u|∂Ω0, u−→μ >0 as|x| −→ ∞,
DL
whereΩ RN\ωis an exterior domain inRN, ω ⊂ RNis a bounded domain with smooth boundary, andN >2, q >1.Among other results, they prove under the assumption that
K1K ∈CαlocΩ, K ≥ 0, K /≡0,and there existC, , M > 0 such that|Kx| ≤ C|x|−lfor
|x| ≥Mwithl≥2
that there existsμ∗>0 such thatDLhas at least one solution forμ∈0, μ∗and no solution forμ∈μ∗,∞.Furthermore, ifK∈L1Ω,then the solution atμμ∗exists and is unique.
We expect that problemDLmay have certain bifurcation phenomenon of solutions with respect toμso that there should be at least one more solution forμ∈0, μ∗.This is our goal for this paper and our first result comes out as follows. Assumeq > p−1 and
Kthere existsβ > p−1 such that∞
r0rβKrdr <∞.
Then, there exist μ0 ≥ μ∗ > 0 such thatP Di, i 1,2,has at least two positive radial solutions forμ∈ 0, μ∗,at least one positive radial solution forμ ∈μ∗, μ0and no positive radial solution forμ∈μ0,∞.
We notice that this result is partial since the existence of multiple solutions on interval μ∗, μ0is not obvious. This is mainly caused by coarse topological structure of solution space.
If indefinite weightK|x|is of the form|x|−lwithl > p, then we can proveμ∗μ0in the above conclusion for problemP D1, that is, there existsμ∗>0 such thatP D1has at least two, one, or no positive solutions according toμ∈0, μ∗, μμ∗, orμ∈μ∗,∞,respectively.
This is our second result for this paper. For proofs, we employ global continuation theorem and fixed point index theory based on a weighted space as the solution space.
It is interesting to see whether the exponentlpis critical or not in the sense of existence of positive radial solutions. We end by answering this question that ifl≤p,then problemP D1does not have a positive radial solution.
Questions for global results or critical sense of exponent for existence of problemP D2are not answered in this work, so we leave them to the readers. A partial answer to the question for the nonexistence results to problemP D2is known in2,3.
This paper is organized as follows. In Section 2, we introduce well-known theorems such as the global continuation theorem, the generalized Picone identity, and a fixed point index theorem for the index computation. InSection 3, we introduce several transformations to obtain equivalent one-dimensional p-Laplacian problems and also prove the existence of unbounded continuum of positive solutions using the global continuation theorem. In Section 4, figuring the shape of the unbounded continuum inSection 3, we get the existence, nonexistence, and multiplicity of solutions introduced as the partial result. In Section 5, introducing weighted spaces, we improve the result inSection 4to a global one. InSection 6, we prove a nonexistence result which gives, in some sense, a critical exponent of existence and nonexistence.
2. Preliminaries
In this section, we give some known theorems which will be used in the following sections.
Theorem 2.1see4, the global continuation theorem. LetXbe a Banach space andKan order cone inX.Consider
xHμ, x, 2.1
whereμ∈Randx∈K.IfH:R×K → Kis completely continuous andH0, x 0 for allx∈K.
ThenCK,the component of the solution set of 2.1containing0,0, is unbounded.
Theorem 2.2see5, the generalized Picone identity. Define
lpy ϕp
yb1tϕpy,
Lpz ϕp
zb2tϕpz. 2.2
Ifyandzare any functions such thaty, z, b1ϕpy, b2ϕpzare differentiable onIandzt/0 for t∈I,the generalized Picone identity can be written as
d dt
|y|pϕp
z ϕpz −yϕp
y
b1−b2
|y|p−
|y|p p−1 yz
z
p−pϕpyyϕp z z
−ylpy |y|p ϕpzLpz.
2.3 Remark 2.3. By Young’s inequality, we get
|y|p p−1 yz
z
p−pϕpyϕp z z
≥0, 2.4
and the equality holds if and only if sgnysgnzand|y/y|p|z/z|p.
Theorem 2.4see6. LetXbe a Banach space,Ka cone inX, andObounded open inX.Let 0∈ O andA:K∩ O → Kbe condensing. Suppose thatAx /νxfor allx∈K∩∂Oand allν≥ 1.Then, iA, K∩ O, K 1.
3. The existence of unbounded continuum
In this section, we introduce several transformations to obtain one-dimensionalp-Laplacian problems which we will mainly analyze and then we prove the existence of unbounded continuum of positive solutions of the problem using the global continuation theorem. By consecutive changes of variables,r |x|, ur u|x|, andt r/r0−N−p/p−1, zt ur, problemP D1is equivalently written as
ϕpzhtzq 0, t∈0,1,
z0 μ >0, z1 0, 3.1
whereϕps |s|p−2s, p >1,andhis given by
ht p−1 N−p
p
r0pt−pN−1/N−pK
r0t−p−1/N−p
. 3.2
We notice thathis singular att0 and by conditionK, hsatisfies H11
0sγhsds <∞,for someγ < p−1.
For more general consideration, we assume that the coefficient functionhmay be singular at t0 and/or 1 which satisfies
H1/2
0 ϕ−1p 1/2
s hτdτds1
1/2ϕ−1p s
1/2hτdτds <∞.
Obviously, we see that conditionH1implies conditionH.Introducingut zt/μ,we can rewrite problem3.1as
ϕpuλhtuq0, t∈0,1,
u0 1, u1 0, 3.3
whereλμq−p1.Problems3.1and3.3share the same bifurcation phenomena with respect toμandλ,respectively.
Similarly, if we use transformationt1−r/r0−N−p/p−1,then problemP D2is also written as3.1withhgiven by
ht p−1 N−p
p
r0p1−t−pN−1/N−pK
r01−t−p−1/N−p
. 3.4
Notice thathis singular att1 and by the conditionK, hsatisfies H21
01−sγhsds <∞,for someγ < p−1.
We see that conditionH2implies conditionH,and thus, for radial problem P D2, it is also enough to consider problem 3.3 withhsatisfying H.Since both problems P Di, i 1,2,can be transformed to the form3.3, we will mainly consider problemPλ given as follows for more general arguments:
ϕp
ut
λhtfut 0, t∈0,1,
u0 a >0, u1 0, Pλ
whereλis a positive real parameter andf∈CR,RwithR 0,∞. h∈C0,1,Rmay be singular att0 and/ort1.Let us assume the following condition:
F1fu>0,for allu >0.
To fulfill conditions in the global continuation theorem, we need to consider problems with Dirichlet boundary condition. For this, we substitutevt ut−a1−tto get the following equivalent problem:
ϕp
vt−aλhtfvt a1−t 0, t∈0,1,
v0 0v1. Pλ
DenoteK{u∈C00,1:uis concave}.Then, it is easy to see thatKis an order cone. Let us define operatorH:R×K → C0,1as follows:
Hλ, vt
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩ t
0
ϕ−1p Aλ,v
s
λhτfvτ a1−τdτ−ϕpa
dsat, if 0≤t≤Aλ,v, 1
t
ϕ−1p s
Aλ,v
λhτfvτ a1−τdτϕpa
ds−a1−t, ifAλ,v≤t≤1, 3.5 where
Aλ,v
0
ϕ−1p Aλ,v
s
λhτfvτ a1−τdτ−ϕpa
dsaAλ,v
1
Aλ,v
ϕ−1p s
Aλ,v
λhτfvτ a1−τdτϕpa
ds−a1−Aλ,v.
3.6
Then by conditionHand the definition ofAλ,v,we can easily see thatHis well defined and HR×K⊂K.Furthermore,uis a positive solution ofPλif and only ifuHλ, uonK.
We can easily see thatH is completely continuous onR×K.The proof basically follows on the lines of Lemmas 2 and 3 in7. SinceH0, u 0 for allu∈KandHλ,0/0 forλ >0,as an application ofTheorem 2.1, we have unbounded continuum of solutions as follows.
Theorem 3.1. Assume that H and F1 hold. Then, there exists an unbounded continuum C bifurcating from0,0in the closure of the set of positive solutions ofPλinR×K.
Corollary 3.2. Assume that H and F1 hold. Then, there exists an unbounded continuum C bifurcating from0, u0,whereu0t a1−t,in the closure of the set of positive solutions of Pλin R×K.
4. The shape of continuum
In this section, we will figure the shape of unbounded subcontinuumCof positive solutions of problemPλknown to exist byCorollary 3.2:
ϕp
utλhtfut 0, t∈0,1,
u0 a >0, u1 0, Pλ
wheref∈CR,R, h∈C0,1,R.We assume an additional condition for this section:
F2f∞limu→ ∞fu/up−1 ∞.
Using the generalized Picone identity and the properties of thep-sine function8,9, we obtain the following lemma.
Lemma 4.1. Assume thatF1andF2hold. Letube a positive solution of Pλ. Then, there exists λ >0 such thatλ≤λ.
Proof. Sinceuis concave andu0 a, ut≥1/4a, for allt∈0,3/4, then it follows from F1andF2that there existsb >0 such thatfu> bup−1, foru≥1/4a.This implies
0ϕp
utλhtfut> ϕp
ut
λbhtϕput, t∈ 0,3 4
. 4.1
Puttingm:mint∈1/4,3/4 ht>0,we have ϕp
utλbmϕput<0, t∈ 1 4,3
4
. 4.2
It is easy to check thatwt Sq2πpt−1/4is a solution of ϕp
wt 2πp
p
ϕpwt 0, t∈ 1 4,3
4
,
w 1 4
0w 3 4
,
4.3
whereSqis theq-sine function with 1/p1/q1 andπp2πp−11/p/psinπ/p.Taking yw, zu, b1 2πpp, andb2λbmin2.3and integrating from 1/4 to 3/4,we have
3/4
1/4
2πp
p−λbm
|w|pdt≥0. 4.4
Thus,
λ≤2πpp
bm :λ. 4.5
Lemma 4.2. Assume thatF2holds. LetJ be a compact interval in0,∞.Then, for allλ ∈J,there existsMJ>0 such that all possible positive solutionsuofPλsatisfyu∞≤MJ.
Proof. Suppose on the contrary that there exists a sequenceunof positive solutions ofPλn withλn⊂Jα, βandun∞ → ∞asn → ∞.It follows from the concavity ofunthat
unt≥1
4un∞, 4.6
for allt ∈1/4,3/4.TakeM 22πpp/αmwithm mint∈1/4,3/4ht >0.ByF2,there existsK >0 such thatfu> Mϕpu,for allu > K.From the assumption, we getuN∞>4K for sufficiently largeN.Therefore by4.6, we have
f uNt
> Mϕp
uNt
, t∈ 1 4,3
4
. 4.7
Hence, we have
ϕp
uNtαMmϕp
uNt
<0, t∈ 1 4,3
4
. 4.8
As in the proof ofLemma 4.1, forwt Sq2πpt−1/4,takingyw, zu, b1 2πpp, andb2αMmin2.3, we obtain
M≤ 2πp
p
αm . 4.9
This is a contradiction.
We now state and prove the main theorem in this section.
Theorem 4.3. Assume thatH,F1, andF2hold. Then, there exist 0< λ∗≤λ∗such thatPλhas two positive solutions for 0< λ < λ∗,one positive solution forλ∗≤λ≤λ∗,and no positive solution for λ > λ∗.
Proof. Defineλ∗ :sup{μ: problemPλhas at least two positive solutions for allλ∈0, μ}.
Then by Lemmas4.1and4.2,λ∗ < ∞.Suppose that there existsλ ≥ λ∗ such thatPλhas a positive solution, sayu, that is,
ϕp
ut−a
λhtf
ut a1−t
0, t∈0,1,
u0 0u1. 4.10
For fixedλ∈0,λ, defineTλ:C0,1 → C0,1by
Tλut
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩ t
0
ϕ−1p B
s
λhτfγuτ−a1−τdτ−ϕpa
dsat, 0≤t≤B, 1
t
ϕ−1p s
B
λhτfγuτ−a1−τdτϕpa
ds−a1−t, B≤t≤1, 4.11 whereγ :R → Ris defined by
γu
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
ut, ifu >ut, u, if 0≤u≤ut, 0, ifu <0,
4.12
andBsatisfies B
0
ϕ−1p B
s
λhτfγuτ−a1−τdτ−ϕpa
dsaB
1
B
ϕ−1p s
B
λhτfγuτ−a1−τdτϕpa
ds−a1−B.
4.13
It is easy to check thatTλ is completely continuous onC0.1.Let us consider the following modified problem:
ϕp
ut−aλhtfγut a1−t 0, t∈0,1,
u0 0u1. Mλ
Then, solutionuof Mλis concave and nontrivial. It follows from the definition ofγ and the continuity offthat there existsR1 >0 such thatTλu∞< R1for allu∈C0,1.Then, by
Schauder’s fixed point theorem there existsu∈C0,1such thatTλuu.Hence,uis a positive solution ofMλ.
We claim thatut ≤ ut, for allt ∈ 0,1.If the claim is not true, then there exists an intervalt1, t2⊂0,1such thatut1 ut 1, ut2 ut 2,andut>ut, for allt∈t1, t2. Putαt ut−ut. Then,αt1 0αt2and there exists an intervalb, c⊂t1δ, t2−δ such thatαb ub−ub > 0 andαc uc−uc < 0 for sufficiently smallδ > 0.
Therefore, we have
ub−a >ub−a, uc−a <uc−a. 4.14 It follows from the monotonicity ofϕpthat
ϕp
ub−a
> ϕp
ub−a
, ϕp
uc−a
< ϕp
uc−a
. 4.15
Since 0< λ <λ, we get 0> ϕp
uc−a
−ϕp
uc−a
−ϕp
ub−a ϕp
ub−a
ϕp
uc−a
−ϕp
ub−a
− ϕp
uc−a
−ϕp
ub−a
c
b
ϕp
ut−a− ϕp
ut−a dt
c
b
−λhtfγut a1−t λhtf
ut a1−t dt
λ−λc
b
htf
ut a1−t dt >0.
4.16
This contradiction impliesut≤ut, for allt∈0,1.Therefore, by the definition ofγ, uturns out a positive solution ofPλ. Defineλ∗ sup{λ :Pλhas at least one positive solution}.
Then byLemma 4.1,λ∗ <∞.Furthermore byLemma 4.2and compactness ofH,we can show thatPλ∗has a positive solution in frame of standard limit argument and this completes the proof.
Take fu uq,q > p −1,in problemPλ, then by the transformation arguments in Section 3, solutions ofPλcorrespond to those of problem3.1which is the radial problem of P Di, i1,2.In this case, conditionsFi,i 1,2, inTheorem 4.3are redundant and we get the following corollary.
Corollary 4.4. Assumeq > p−1 and assume that Kthere existsβ > p−1 such that∞
r0 rβKrdr <∞.
Then, there existμ0 ≥μ∗> 0 such thatP Di, i 1,2,has at least two positive radial solutions forμ∈0, μ∗,at least one positive radial solution forμ∈μ∗, μ0,and no positive radial solution for μ∈μ0,∞.
5. Global existence result
Multiplicity of solutions onλ∗, λ∗is not known inTheorem 4.3. Analytic difficulty on this range is caused by lack of topological properties in solution space C0,1, mainly lack of controllability of derivatives of solutions at the boundary. In this section, we overcome this difficulty by employing a weighted space as new solution space specially for problemP D1. For this purpose, let us consider the followingp-Laplacian problem:
div
|∇u|p−2∇u
|x|−luq0 in|x|> r0, E
u|∂Ω0, u−→μ >0 as|x| −→ ∞. D1
We notice thatK|x| |x|−l withl > psatisfies conditionK.By transformationsr
|x|, ur u|x|,andt r/r0−N−p/p−1, ur zt,we obtain ϕp
zt p−1 N−p
p
r0pt−pN−1p−1l/N−pzqt 0, t∈0,1, z0 μ >0, z1 0.
5.1
For α pN −1−p−1l/N−p,condition l > pcorresponds to α < p. By another transformationut zt/μ,the above problem can be transformed into
ϕp
utλt−αuqt 0, t∈0,1,
u0 1>0, u1 0, 5.2
whereλμq−p1.
As inSection 3, we consider problemEλgiven as follows for more general arguments:
ϕp
utλt−αfut 0, t∈0,1,
u0 a >0, u1 0, Eλ
whereλis a positive real parameter andf ∈CR,R.We give an additional assumption in this section:
F3fis nondecreasing.
Now the aim of our work here is to investigate bifurcation phenomena of positive solutions for problemEλ. Again introducing vt ut−a1−t,we rewriteEλto the following equivalent Dirichlet boundary problem:
ϕp
vt−aλt−αfvt a1−t 0, t∈0,1,
v0 0v1. Eλ
We first state the main theorem in this section.
Theorem 5.1. Assume thatF1,F2,andF3hold. Also assume thatα < p.Then, there existsλ∗>0 such thatEλhas at least two positive solutions forλ∈0, λ∗,at least one positive solution forλλ∗, and no positive solution forλ∈λ∗,∞.
Ifα <1,thenht t−αis of classL10,1.Thus, solution space isC10,1and by typical Leray-Schauder degree argument in the frame ofC1-topology, we can prove that the theorem is truesee,10. Therefore, in this section, we focus on the case 1≤α < p.Define
wt
⎧⎨
⎩
tα−1/p−1, if 1< α < p, min
−lnt−1/p−1,1
, ifα1. 5.3
We can easily know that
tlim→0wt 0, 0< wt≤1, t∈0,1, w−1∈L10,1. 5.4 The following lemma is essential to introduce our weighted spaceCw0,1and very useful to construct a bounded open set of solutions in the space for fixed point index computation.
Lemma 5.2. Ifuis a solution ofEλ, thenwu∈C0,1and 0< lim
t→0wut<∞. 5.5
Proof. Letube a solution ofEλ. Then, we have ut ϕ−1p
λ
A
t
τ−αfuτ a1−τdτ−ϕpa
a, 5.6
whereuA 0.Sincew∈C0,1andu∈C0,1,we only need to show 0< lim
t→0wut<∞. 5.7
In fact, if 1< α < p,then by L’Hospital’s rule and limt→0wt 0,we have
tlim→0wut lim
t→0
ϕ−1p
λA
t τ−αfuτ a1−τdτ−ϕpa t1−α
awt
ϕ−1p λ 1 α−1fa
>0.
5.8
Ifα1,then similarly we may obtain
tlim→0
wu
t ϕ−1p λfa>0. 5.9 Thus, the proof is complete.
DefineCw0,1 {u ∈ C00,1∩C10,1 : limt→0wutexists}.We notice that if u∈Cw0,1,then there exists an extensionwu∈C0,1ofwusuch that
wut
⎧⎨
⎩
tlim→0
wu
t, t0, wu
t, t∈0,1. 5.10 Define
uwu∞wu∞. 5.11
Proposition 5.3. Cw0,1, · wis a Banach space.
Proof. It is easy to see thatCw0,1is a normed linear space. We only need to check thatCw0,1 is complete. In fact, letunbe a Cauchy sequence inCw0,1.That is,unis a Cauchy sequence inC00,1andwunis a Cauchy sequence inC0,1.Since bothC00,1andC0,1are Banach spaces, there existu∈C00,1andv∈C0,1such thatun → uinC00,1andwun → vin C0,1.Sincewt>0,fort∈0,1,there existsv1 ∈C0,1such thatvt wtv1t,for all t∈0,1.Forδ >0,we knowun → v1inCδ,1.This impliesv1 ≡uinδ,1.Sinceδ >0 is arbitrary,wun → wupointwise int∈0,1.Therefore, by the uniqueness of limit,wu≡von 0,1.Sincewun → vinC0,1,we have
v0 lim
n→ ∞wun0 lim
n→ ∞lim
t→0wunt lim
t→0 lim
n→ ∞wunt lim
t→0wut.
5.12
Therefore,u∈Cw0,1andwu≡von0,1.This impliesun → uinCw0,1and the proof is complete.
LetK{u∈Cw0,1|uis concave on0,1}.Then, it is easy to check thatKis an order cone. DefineH :R×K → Cw0,1by
Hλ, ut
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩ t
0
ϕ−1p Aλ,u
s
λτ−αfuτ a1−τdτ−ϕpa
dsat, 0≤t≤Aλ,u, 1
t
ϕ−1p s
Aλ,u
λτ−αfuτ a1−τdτϕpa
ds−a1−t, Aλ,u≤t≤1, 5.13
where Aλ,u
0
ϕ−1p Aλ,u
s
λτ−αfuτ a1−τdτ−ϕpa
dsaAλ,u
1
Aλ,u
ϕ−1p s
Aλ,u
λτ−αfuτ a1−τdτϕpa
ds−a 1−Aλ,u
.
5.14
Assume that F1 holds. Then, by the similar argument in the proof of Lemma 5.2 and the definition ofH,we see thatHis well defined andHR×K⊂K.Furthermore,uis a positive solution of Eλ if and only if u Hλ, u on K.The following lemma can be proved by standard argument and we skip the proof.
Lemma 5.4. Assume thatF1holds. Then,His completely continuous onR×K.
SinceH0, u 0,for allu ∈ K andHλ,0/0,forλ > 0,Lemma 5.4and the global continuation theorem imply that there exists an unbounded continuumCof positive solutions ofEλbifurcating from0,0.
Lemma 5.5. Assume thatF1andF2hold. Letube a positive solution of Eλ. Then, there exists λ >0 such thatλ≤λ.
Proof. Takinght t−αwithα < p,we can prove this lemma by the same argument in the proof ofLemma 4.1.
Lemma 5.6. Assume thatF2holds and letIbe a compact interval in0,∞.Then, there existsbI>0 such that for all possible positive solutionuofEλwithλ∈I,one has
uw≤bI. 5.15
Proof. Assume on the contrary that there exists a sequenceunof a positive solution ofEλn with λn ∈ I such that unw → ∞as n → ∞.We claim un∞ → ∞ asn → ∞. This contradicts Lemma 4.2 and the proof is complete. If the claim is not true, then there exists M1>0 such thatun∞≤M1,for alln.Sinceunis a positive solution ofEλn,we get
−ϕp
unt−a
λnt−αf
unt a1−t
≤dt−α, t∈0,1, 5.16 wheredsupI maxu∈0,M1afu.Integrating this fromtto 1, we have
ϕp
un1−a
−ϕp
unt−a≤d 1
t
s−αds, ϕp
unt−a≤ϕp
un1−ad 1
t
s−αds.
5.17
Sinceϕ−1p is increasing, we obtain unt−a≤ϕ−1p
ϕp
un1−ad 1
t
s−αds
,
≤22−p/p−1
un1aϕ−1p
d
1
t s−αds
.
5.18
Therefore, by the similar computation in the proof ofLemma 5.4, we obtain wtunt≤22−p/p−1
un1
12p−2/p−1 aϕ−1p
dwp−1t 1
t
s−αds
≤22−p/p−1
Y1
12p−2/p−1 aϕ−1p
d max
t∈0,1
wtp−1
1
t
s−αds
<∞,
5.19
where
Yt
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩ t
0
ϕ−1p
d B
s
τ−αdτ−ϕpa
dsat, 0≤t≤B, 1
t
ϕ−1p
d s
B
τ−αdτϕpa
ds−a1−t, B≤t≤1,
5.20
withYB Y∞.This implies that{wun∞}is bounded. This contradicts our assumption unw → ∞and the claim is proved.
Let us assume that problemEλhas a positive solution atλ∗>0 so letu∗be a positive solution ofEλ∗.We see thatu∗satisfies
ϕp
u∗t−aλ∗t−αfu∗t a1−t
0, t∈0,1. 5.21 Consider a fixed parameterλ∈0, λ∗.ForN >0,let us define
ΩN
u∈Cw0,1|0< ut< u∗t, t∈0,1, 0< wu 0
< wu∗ 0
, u∗
1−
< u 1−
<0 andwu∞< N .
5.22
Then byLemma 5.2,ΩN is bounded and open in Cw0,1.Consider the following modified problem:
ϕp
ut−aλt−αfγut a1−t 0,
u0 0u1, Mλ1
whereγ :R → Rby
γu
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
u∗t, ifu > u∗t, u, if 0≤u≤u∗t, 0, ifu <0.
5.23
Lemma 5.7. Assume thatF1,F2,andF3hold. Ifuis a positive solution ofMλ1forλ∈0, λ∗, thenu∈ΩN∩K,for someN >0.
Proof. Letube a positive solution ofMλ1. We first show 0 < ut ≤ u∗t, t ∈ 0,1.If it is not true, there existst1, t2 ⊂ 0,1such thatut > u∗tfort ∈ t1, t2, ut1 u∗t1,and ut2 u∗t2.Sinceu−u∗∈C0t1, t2,there existsA∈t1, t2such that
uA u∗A, uA−u∗A>0. 5.24
Sinceγut a1−t≤u∗t a1−t, λ < λ∗, andfis nondecreasing, we have
λ∗fu∗t a1−t> λfγut a1−t. 5.25
This implies
ϕp
ut−a
λ∗t−αfu∗t a1−t>0, t∈0,1. 5.26 From5.21and5.26, we have
ϕp
u∗t−a−ϕp
ut−a
<0, t∈0,1. 5.27
Fort∈A,1,integrating5.27fromAtot, we haveu∗t≤ut.Again, integrating this from Ato 1, we get
u∗A≥uA. 5.28
This contradicts5.24.
Second, we show ut < u∗t, t ∈ 0,1.If it is not true, then by the first argument and5.27, we have only one case; there exist t3 ∈0,1andδ1 >0 such that ut3 u∗t3, ut< u∗t,t∈t3−δ1, t3δ1\ {t3}, andut3 u∗t3.Fort∈t3−δ1, t3,integrating5.27 fromttot3,we have
u∗t≥ut, fort∈
t3−δ1, t3
. 5.29
Again integrating this fromt3−δ1/2 tot3,we get u∗ t3−δ1
2
≤u t3−δ1
2
5.30
and this is a contradiction.
Third, we show that 0 < limt→0wut < limt→0wu∗t. By the second argument, γut ut,fort∈0,1.By the similar calculation as inLemma 5.2, we have
tlim→0wut ϕ−1p λfa< ϕ−1p λ∗fa lim
t→0wu∗t, 5.31
ifα1.The case 1< α < pis similar.
Fourth, we show that 0> u1> u∗1.We first claim that there existsc∈0,1such that uc> u∗c.Indeed, otherwise,ut≤u∗t,for allt∈0,1.Integrating this fromtto 1, we have
ut≥u∗t, fort∈0,1. 5.32
This is a contradiction by the second argument. Integrating5.27fromcto 1, we obtain ϕp
u∗1−a
−ϕp
u1−a
< ϕp
u∗c−a
−ϕp
uc−a
<0 5.33
and thus
u∗1< u1. 5.34
Sinceuis a positive solution ofMλ1, obviouslyu1<0.
Finally, we showwu∞< Nfor someN >0.Since 0≤ut≤u∗tfor allt∈0,1,by the similar calculation in the proof ofLemma 5.6, we have
wut≤22−p/p−1
Y1
2p−2/p−11 aϕ−1p
f∗ max
t∈0,1
wtp−1
1
t s−αds
N,
5.35 for allt∈0,1,wheref∗λmaxu∈0,u∗a fu>0,and
Yt
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩ t
0
ϕ−1p
f∗
B
s
τ−αdτ−ϕpa
dsat, 0≤t≤B, 1
t
ϕ−1p
f∗
s
B
τ−αdτϕpa
ds−a1−t, B≤t≤1,
5.36
andBis defined by B
0
ϕ−1p
f∗
B
s
τ−αdτ−ϕpa
dsaB 1
B
ϕ−1p
f∗
s
B
τ−αdτϕpa
ds−a1−B,
5.37 and this completes the proof.
We now prove the main theorem in this section.
Proof ofTheorem 5.1. Letλ∗sup{μ|Eλhave at least two positive solutions for allλ∈0, μ}.
Then, by Lemmas5.5and5.6, 0 < λ∗ ≤ λ.By the choice ofλ∗,Eλhas at least two positive solutions forλ ∈ 0, λ∗and at least one positive solution atλ λ∗.We will show thatEλ has no positive solution for allλ > λ∗.On the contrary, assume that there existsλ∗ > λ∗such thatEλ∗has a positive solution. We claim thatEλhas at least two positive solutions forλ∈ λ∗, λ∗.Then, this contradicts to the definition ofλ∗and the proof is done. DefineM:K → K by
Mut
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩ t
0ϕ−1p
Au
s λτ−αfγuτ a1−τdτ−ϕpa
dsat, 0≤t≤Au, 1
t
ϕ−1p s
Au
λτ−αfγuτ a1−τdτϕpa
ds−a1−t, Au≤t≤1, 5.38 where
Au
0
ϕ−1p Au
s
λτ−αfγuτ a1−τdτ−ϕpa
dsaAu
1
Au
ϕ−1p s
Au
λτ−αfγuτ a1−τdτϕpa
ds−a 1−Au
.
5.39
Then,M:K → Kis completely continuous anduis a solution ofMλ1if and only ifuMu onK.By simple calculation, we can easily check that there existsR1>0 such thatMuw< R1, for allu∈K.TakingR1big enough satisfyingBR1⊃ΩNand applyingTheorem 2.4, we get
i
M, BR1∩K, K
1. 5.40
ByLemma 5.7and the excision property, we get i
M,ΩN∩K, K i
M, BR1∩K, K
1. 5.41
Since problemEλis equivalent to problemMλ1onΩN∩K,we conclude that Eλhas a positive solution inΩN∩K.Assume thatHλ,·has no fixed point in∂ΩN∩Kotherwise, the proof is done!. Then,iHλ,·,ΩN∩K, Kis well defined and by5.41, we have
i
Hλ,·,ΩN∩K, K
1. 5.42
ByLemma 5.5, we may chooseλN0> λsuch thatEλN0has no solution inK.By a priori estimate Lemma 5.6withI λ, λN0,there existsR2> R1such that for all possible positive solutions uofEμwithμ∈λ, λN0,we have
uw< R2. 5.43
Defineh:0,1×BR2∩K → Kby hτ, u H
τλN0 1−τλ, u
. 5.44
Then, by the similar argument asLemma 5.4,his completely continuous on0,1×Kand by Lemma 5.6,hτ, u/u,for allτ, u∈0,1×∂BR2∩K.By the property of homotopy invariance, we have
i
Hλ,·, BR2∩K, K i
H λN0,·
, BR2∩K, K
0. 5.45
By the additive property and5.42, we have i
Hλ,·,
BR2\ΩN
∩K, K
−1. 5.46
ThereforeEλhas another positive solution inBR2\ΩN∩K.This completes the claim.
Corollary 5.8. Assume thatl > pandq > p−1.Then, there existsμ∗>0 such thatE D1has at least two positive radial solutions forμ∈0, μ∗,at least one positive radial solution forμμ∗,and no positive radial solution forμ∈μ∗,∞.
6. Nonexistence
In this section, we will prove the nonexistence result of nonnegative solution of problemEλ ifα≥p.
Theorem 6.1. Assume thatF1holds and also assumeα≥p.Then,Eλhas no positive solution.
Proof. First, we prove the caseα p.It is obvious that problem Eλdoes not have a trivial solution. Suppose on the contrary that there is a positive solutionuofEλwhenαp.Since uis concave, there exists uniquez∈0,1such thatuz a/2.Therefore, we haveut≥a/2, fort∈0, z.This implies
−ϕpu≥λmt−p, 6.1
fort∈0, z,wheremminv∈a/2,ufv>0.Forx∈0, z/2,we can easily know that x1−α−z1−α>
1−21−α
x1−α. 6.2
Integrating6.1fromxtozand by6.2we obtain ϕ−1p
ϕp ux
−ϕp uz
≥Cx−1, 6.3
whereCϕ−1p λm1/p−11−21−p.Sinceϕpux−ϕpuz≥0,we obtain
x−1≤γ1/p−1C−1uxuz, 6.4
whereγq max{1,2q−1}.It is enough to consider the following two cases:ithere existsl ∈ 0, z/2such thatux>0,for allx∈0, landiiux<0 for allx∈0, z/2.For the first case, lety∈0, l,then integrating6.4fromytol,we get
uy≤ul uzl−y γ1/p−1−1 C−lnllny. 6.5
Lettingy → 0,we haveuy → −∞and this is a contradiction tou0 a.For the second case, lety∈0, z/2,then integrating6.4fromytoz/2,
lnz
2 −lny≤γ1/p−1C−1
−u z 2
uy uz z 2−y
. 6.6
This implies
uy≥u z 2
−uz z 2 −y
γ1/p−1−1 C lnz 2−lny
. 6.7
Lettingy → 0,we obtainuy → ∞.Once again, this is a contradiction tou0 a.Similarly, we can prove the same conclusion for the caseα > p.
Remark 6.2. Instead of conditionF1,if we assume that there exists >0 such thatfu>0 for u≥,then the conclusion ofTheorem 6.1still remains true.
Corollary 6.3. Assumel≤p.ThenE D1has no positive radial solution.
One of the referee informed thatCorollary 6.3was proven under more general set up in 2.
Acknowledgments
The authors express their thanks to the referees for their valuable comments and suggestions.
This work was supported by the Korea Research Foundation Grant funded by the Korean GovernmentMOEHRDKRF-2005-070-C00010.
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